 In this problem we have a closed-end cylindrical tube with a diameter of 250 mm and a wall thickness of 5 mm, which is subjected to an internal pressure P equal to 3 MPa and a torque T equal to 15 kN about its axis. I'm using more circle, we need to determine the principal stresses and the maximum shear force and estimate the orientation of the plane at which they occur. Then first we need to calculate what are the stresses created by the pressure and the torque. So if we consider this small element here, we will define this to be the stress in the x-direction, so this is the x-direction, this is the y-direction, sigma y, and this is the shear stress tau xy. So let's determine what are these stresses. We have first the stress due to pressure, and we know from our previous exercise that if we have a cylinder with an internal pressure P, the stress in the longitudinal direction sigma x is equal to P divided by 2T, where T is the thickness of the wall. So this is equal to under stress in the y-direction for this cylinder. We have also the solution from our previous exercise, this is equal to P divided by the wall thickness, then this is equal to... And now we can determine the stress due to torsion. Of course due to torsion we have only shear stress, then we can use the torsional formula. We know that the stress divided by r is equal to the torque divided by the polar moment of inertia, then from here the shear is equal to T r divided by the polar moment of inertia. And since we can use the fin wall approximation, we know that in this case the polar moment of inertia is equal to 2 pi r cubed times T. Then from here we have this is equal to T times r, and this is finally equal to... Then these are the stresses in the x and y-direction and the shear stress. Then now we can start drawing our Mohr circle, this is our small element. Then we can first define the center point, which is equal to the average stress. So this is equal to sigma x plus... First we can start calculating what is the center of the circle. We know that it's equal to sigma average and this is equal to sigma x plus sigma y. Divided by 2, so this is equal to... We know that the radius is equal to sigma x minus the average squared plus the shear stress squared and the square root of this. So this is equal to... Now I can start drawing my Mohr circle, this is my reference. I know that the center is located more or less at 56. So this is sigma average. And I can start defining the plane A, point A. I know that the shear stress here is creating a counterclockwise rotation. So according to the criteria that we are using, this is a negative shear stress. So negative 36, 30.56. And the corresponding sigma x is 37.5. So the point is more or less here, is 0.1. And 0.2 of course, the rotation is clockwise, so positive 30.56. And the corresponding normal stress is equal to 75 megapascals. So more or less around here. Then this is... These are my planes, A and B. This is sigma x and sigma y. And now I can finish drawing my circle more or less like this. So these are the planes of the principal stresses. This is sigma 1. And this is the second principal stress. So now we need to calculate them. We know that the principal stress 1 is equal to... And sigma 2 is equal to... Right? Remember that I didn't say this, but sigma 1 is the average. Plus radius and sigma 2 is the average. Minus the radius. So from here now, we know, because the problem is asking for this, what is the maximum shear stress, which is equal to the radius. Then the maximum shear stress is equal to 38, 35.85 megapascals. We have our principal stresses. And we need to finally determine what is the orientation of the principal planes with respect to the reference that we have right now. So for example, with respect to the plane A, we know that the angle of rotation in the Mohr circle is 2 times the angle of rotation in the real problem. So this is the angle 2 phi 1 that the principal axis forms with A. And of course, this is the angle 2 phi 2 that this second principal axis is forming with the plane A. So from the geometry of the problem, we can measure that 2 phi 1 is more or less 135 degrees. So from here, we know that phi 1 is equal to 7.5 degrees counterclockwise. From A, they have the same direction of rotation. And 2 phi 2, which is equal to 2 phi 1 plus 180. So this is equal to 315. So from here, we know that phi 2 is equal to 157.5 degrees. Again, counterclockwise from A.